The inequality given is:

$\frac{(x - 7)(x + 5)}{x} \geq 0$

Let's solve this algebraically by following these steps:

Step 1: Identify critical points

The expression involves factors $(x - 7)$, $(x + 5)$, and $x$. The critical points are the values of $x$ that make any factor equal to zero:

1. $x = 0$ (from the denominator).
2. $x = 7$ (from $(x - 7) = 0$).
3. $x = -5$ (from $(x + 5) = 0$).

So, the critical points are $x = -5$, $x = 0$, and $x = 7$.

Step 2: Determine the sign in each interval

We now test the sign of the expression in the intervals determined by the critical points: $(- \infty, -5)$, $(-5, 0)$, $(0, 7)$, and $(7, \infty)$.

• For $x \in (-\infty, -5)$: Choose $x = -6$: $\frac{(-6 - 7)(-6 + 5)}{-6} = \frac{(-13)(-1)}{-6} = \frac{13}{6}, \quad \text{positive}.$

• For $x \in (-5, 0)$: Choose $x = -1$: $\frac{(-1 - 7)(-1 + 5)}{-1} = \frac{(-8)(4)}{-1} = 32, \quad \text{positive}.$

• For $x \in (0, 7)$: Choose $x = 1$: $\frac{(1 - 7)(1 + 5)}{1} = \frac{(-6)(6)}{1} = -36, \quad \text{negative}.$

• For $x \in (7, \infty)$: Choose $x = 8$: $\frac{(8 - 7)(8 + 5)}{8} = \frac{(1)(13)}{8} = \frac{13}{8}, \quad \text{positive}.$

Step 3: Analyze the critical points

• At $x = -5$, the numerator becomes zero, so $\frac{(x - 7)(x + 5)}{x} = 0$. This satisfies the inequality.
• At $x = 0$, the expression is undefined (division by zero).
• At $x = 7$, the numerator becomes zero, so the expression equals zero, satisfying the inequality.

Step 4: Conclusion

The solution to the inequality is where the expression is either positive or zero. Based on the sign analysis and critical points, the solution in interval notation is:

$(-\infty, -5] \cup (0, 7]$

Would you like a more detailed explanation on any step or have further questions?

Here are 5 related questions for deeper understanding:

1. How would the solution change if the inequality were strict (i.e., $> 0$ instead of $\geq 0$)?
2. What happens if one of the critical points were a repeated root?
3. How can we visualize this inequality on a number line?
4. Why is $x = 0$ excluded from the solution?
5. How do rational inequalities differ from polynomial inequalities?

Tip: Always check for undefined points in rational inequalities to avoid errors in your solution!